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If $\alpha$ represent the square of the distance between the origin and the point of intersection of the lines $x^2-y^2-x+3 y-2=0$ and $\beta$ represent the product of the perpendicular distances from the origin on the pair of lines, then $\alpha \beta=$
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Verified Answer
The correct answer is:
$\left(\frac{5}{2}\right)^{0.5}$
We have, pair of straight line is
$\begin{gathered}
x^2-y^2-x+3 y-2=0 \\
(x-y+1)(x+y-2)=0 \\
x-y+1=0 \text { and } x+y-2=0
\end{gathered}$
Solving these equation, we get
$x=\frac{1}{2} \text { and } y=\frac{3}{2}$
$\therefore$ Intersection point $P\left(\frac{1}{2}, \frac{3}{2}\right)$
$\alpha=O P=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^2}=\frac{\sqrt{10}}{2}$
Perpendicular distance from origin to line $x-y+1=0$
and $x+y-2=0$ are $\frac{1}{\sqrt{2}}$ and $\frac{2}{\sqrt{2}}$ respectively,
$\beta=\frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{2}}=1 \Rightarrow \alpha \beta=\frac{\sqrt{10}}{2} \times 1=\sqrt{\frac{5}{2}}$
${ }^{(*)}$ No option is matched.
$\begin{gathered}
x^2-y^2-x+3 y-2=0 \\
(x-y+1)(x+y-2)=0 \\
x-y+1=0 \text { and } x+y-2=0
\end{gathered}$
Solving these equation, we get
$x=\frac{1}{2} \text { and } y=\frac{3}{2}$
$\therefore$ Intersection point $P\left(\frac{1}{2}, \frac{3}{2}\right)$
$\alpha=O P=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^2}=\frac{\sqrt{10}}{2}$
Perpendicular distance from origin to line $x-y+1=0$
and $x+y-2=0$ are $\frac{1}{\sqrt{2}}$ and $\frac{2}{\sqrt{2}}$ respectively,
$\beta=\frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{2}}=1 \Rightarrow \alpha \beta=\frac{\sqrt{10}}{2} \times 1=\sqrt{\frac{5}{2}}$
${ }^{(*)}$ No option is matched.
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